Resolvability: For all collections of
votes for which two or more alternatives
have a chance of being elected, there must exist some admissible vote
such
that,
if this vote
were added to the collection of votes, one alternative would be elected
with certainty and the rest defeated with certainty. In particular,
the one which
would be elected with certainty with this vote added must be one of those
having
a non-zero chance of being elected without this vote added.

Reasonable
Determinism: For all collections of
votes such that no alternatives
tie pairwise and no majorities are the
same size, there must exist an alternative
which is elected with
certainty.

These two criteria are closely related, satisfaction implying the voting rule
will usually
leave nothing to chance when electing the winner. Compliance with these criteria
means that in large public elections the
outcome will nearly always be deterministic.

It would be undesirable to require complete determinism in all
scenarios. That would
require sacrificing either neutrality of the alternatives or anonymity of the
voters.
To illustrate this, suppose there are two alternatives x
and y, and half of the voters
rank x over y and the other half rank y over x.
Neutrality requires neither x nor y
be privileged over the other, and anonymity requires no voter be
privileged over
any other voters. Therefore in this example both alternatives must be given an equal
chance (½) of being elected. (If one of the alternatives is "continue the status quo,"
it would be
reasonable to violate neutrality by privileging the status quo for
the sake
of stability, rather than
resort to chance when breaking a tie.) If the number of voters
is large, it is unlikely
that exactly half will rank x over y (or y over x). What
matters
more is that the voting rule be deterministic in more likely scenarios, which raises
the question of how to express such a
criterion.

Resolvability is more widely recognized than reasonable
determinism by the political
science community as a desirable criterion. Reasonable
determinism is tailored in
favor of
"pairwise" voting procedures such as MAM, Simpson, Copeland,
etc. (but
not Borda nor Instant Runoff) and
thus should only be used when comparing pairwise
election procedures. Reasonable determinism is mentioned here because MAM
complies, in case anyone wonders and because understanding the proof
(provided
elsewhere) may augment one's understanding of MAM.

Two examples illustrate resolvability and reasonable determinism:

Example 1: Suppose there are 3
alternatives x, y and z. Suppose there are 15
voters
who rank the alternatives as follows:

4

5

6

x

y

z

y

z

x

z

x

y

It can be seen that a majority (9 voters) rank y over z, and
a majority (10 voters) rank x
over y, and a majority (11 voters) rank z over x.
Thus no two majorities are the same size
and no two alternatives are tied pairwise, so reasonable determinism demands
that the
voting procedure elect one of the alternatives with certainty. Consider
the Copeland
procedure, which awards each alternative a point for each pairing it wins plus
a half point
for each pairing it ties, and then chooses the alternative(s) having the
highest score. (This
is the same procedure used to rank teams in many professional sports, such as
baseball.)
In this example, Copeland awards each alternative 1 point. (For instance,
since a majority
ranked x over y and a majority ranked z over x, x
wins the "x versus y" pairing and loses
the "x versus z" pairing; thus x has a score of
1 + 0 = 1.) Therefore Copeland chooses
all three alternatives, proving Copeland does not satisfy reasonable
determinism.
It can also be checked that there is no admissible vote (a ranking of the
alternatives)
which, if added to the 15 votes, would affect the three Copeland scores. Thus
Copeland also does not satisfy resolvability.

Example 2: (Illustrates satisfaction
of resolvability) Suppose there are 3 alternatives x, y and z. Suppose there are 14 voters who rank
the alternatives as follows:

4

5

5

x

y

z

y

z

x

z

x

y

It can be seen that a majority (9 voters) rank y over z, and a
majority (9 voters) rank x
over y, and a majority (10 voters) rank z over x.
Consider the operation of MAM in
this example: Only two of the three majorities can be affirmed since the
three
together
are not internally consistent. (In other words, they cycle.) First MAM affirms
"z over x"
since it has the largest majority. Since both of the other majorities are
the same size,
the one given greater precedence by MAM depends on a tiebreaking strict ranking
constructed using the Random Voter Hierarchy procedure. Thus there is a 4/14
chance that the tiebreaking ranking will be "x over y over z" and a 5/14
chance that
the tiebreaking ranking will be "y over z over x" and a 5/14 chance
that the
tiebreaking
ranking will be "z over x over y." The loser of the "y over z" majority is z, and
the loser
of the "x over y" majority is y. Since there is a 9/14 chance
that the tiebreaking ranking
ranks y over z,
there is a 9/14 chance that MAM gives greater precedence to "y over z"
than to
"x over y." Thus "x over y"
has a 9/14 chance of not being affirmed, and "y over z"
has
a 5/14 chance of not being affirmed. Thus there is a 9/14 chance that MAM
elects y
and a 5/14 chance
that MAM elects z. There is no alternative that MAM elect with
certainty. Now suppose one extra vote that ranks "z over x over y"
were added to the
14 votes. Then, as in example 1, there would be a majority of 9 who rank y over z,
and a majority of 10 who rank x over y, and a majority of 11 who rank z over x.
This
means
MAM would affirm "z over x" and then affirm "x over y" and would not affirm
"y
over z." Thus, if that vote were added, MAM would elect z with certainty and
would defeat x and y
with certainty.

Proof #1 that MAM is resolvable: Assume no alternative in A is
elected with certainty
by MAM
given the votes R. By theorem
"At Least One Top" and inspection of the
definition of MAM, at least one alternative
always
has a non-zero chance of being
elected by MAM. Relabel the alternatives if necessary so that x
denotes an alternative
that
has a non-zero chance of being elected by MAM given R. This means there exists
σ ∈
L(R,A) such that MAM(A,R,σ) = x. We must show there exists an admissible
vote r
such that MAM chooses x with certainty and defeats all other alternatives with
certainty given R+r. Make the following abbreviations:

Proof of 1.4: Since Aff ⊆
Pr and Pr is internally consistent, every
pair in Aff is
consistent with Pr. By 1.2 Aff' ⊆
Pr, so every pair in Aff is consistent with Aff'.
By 1.3,
Aff ⊆ Maj'. Since every pair in Aff is in Maj'
and every pair in Aff is
consistent with Aff', it follows by the definition of Affirmed()
that Aff ⊆ Aff'.
Thus 1.4 has been established. QED

Proof of 1.5: Since x
tops r, Pr contains no pair in which x
is second. Thus,
by 1.2 Aff'
contains no pair in which x is second. It follows by the definition
of Top() that x ∈
Top'. Thus 1.5 has been established.
QED

Case 1.6.2: y ∉
Top. By the definition of Top(),
there
exists a ∈
A
such that (a,y) ∈
Aff. By 1.4 (a,y) ∈
Aff'. Therefore y ∉ Top'.

Since y ∉
Top' in both cases, 1.6 has been established.
QED

By 1.4 and 1.5, Top' = {x}.
Thus MAM'
= x. Since σ' was picked arbitrarily from L(R',A),
it follows that MAM(A,R',σ')
= x for all σ'
∈ O(R',A). This
implies MAM
elect x with certainty and defeats all other alternatives with certainty given R+r.
Thus MAM is resolvable. QED

Proof #2 that MAM is resolvable: We rely on theorem
"MAM2
= MAM" and
show that MAM2 is resolvable. Assume x has a
non-zero chance of being elected
by MAM2. This means there exists σ ∈
L(R,A) and r ∈ L(A)
topped by x such
that r ∈ UndominatedRankings(A,R,σ). Since r is an ordering of A, r is an
admissible vote. We
will show UndominatedRankings(A,R+r,σ')
= {r} for
all σ' ∈
L(R+r,A), after which it will follow easily that MAM2
elects x with certainty
and defeats all other alternatives with certainty given R+r. Abbreviate R' = R+r.
Clearly both of the following statements hold:

(2.1) #R'(a,b)
= 1 + #R(a,b)
for all a,b ∈
A such that r ranks a
over b.
(2.2) #R'(a,b)
= #R(a,b)
for all a,b ∈
A such that r does not rank a
over b.

Pick any σ' ∈
O(R',A) and any r' ∈
L(A) distinct from r.
We aim to show
that r dominates r' given (R',σ').
Make the following abbreviations:

Since r ∈ Undom, r'
does not dominate r given (R,σ).
There are two cases to consider:

Case 2.9: r does not
dominate r' given (R,σ).
Since neither dominates the other
given (R,σ),
Distinct must be empty. This implies #R(a,b)= #R(b,a)
for all a,b ∈
A such that r and r' disagree on the relative
ranking of a and b.
By 2.5 and 2.6, both of the following statements must hold:
(2.9.1) For all a,b
∈ A such that r ranks a
over b
and r' does not rank a
over b, #R'(a,b)= #R'(b,a)
+ 1.
(2.9.2) For all a,b ∈ A such that r' ranks
a over b
and r does not rank a over b, #R'(a,b)= #R'(b,a)
- 1.
By 2.9.1 and 2.9.2, both of the following statements must hold:
(2.9.3) For all a,b ∈ A such that r ranks a
over b
and r' does not rank a
over b, (a,b)
∈ Distinct' ∩
Agreed'(r).
(2.9.4) For all a,b ∈ A such that r' ranks
a over b
and r does not rank a over b, (a,b)
∉ Maj'.
By 2.4, 2.9.3 and 2.9.4, Distinct' ≠ φ
and Distinct' ⊆ Agreed'(r).
This implies LDA' ∈ Agreed'(r).
Thus r dominates r' given (R',σ')
in case 2.9.

Since r dominates r' given (R',σ')
in both cases and since r' was any ranking
distinct from r,
Undom' = {r}. Since σ'
also was picked arbitrarily, this implies UndominatedRankings(A,R',σ')
= {r} for all σ' ∈
L(R',A). Since x
tops r,
MAM2 elects x with certainty and defeats
all other alternatives with certainty
given R'. Thus MAM2
is resolvable. By theorem
"MAM2 = MAM",
MAM is resolvable.
QED